What are the divisors of 5398?

1, 2, 2699, 5398

There is a total of 4 positive divisors.
The sum of these divisors is 8100.
The arithmetic mean is 2025.

2 even divisors

2, 5398

2 odd divisors

1, 2699

How to compute the divisors of 5398?

A number N is said to be divisible by a number M (with M non-zero) if, when we divide N by M, the remainder of the division is zero.

$N mod M = 0$

Brute force algorithm

We could start by using a brute-force method which would involve dividing 5398 by each of the numbers from 1 to 5398 to determine which ones have a remainder equal to 0.

$Remainder = N - (M \times ⌊ \frac{N}{M} ⌋)$

(where $⌊ \frac{N}{M} ⌋$ is the integer part of the quotient)

5398 / 1 = 5398 (the remainder is 0, so 1 is a divisor of 5398)
5398 / 2 = 2699 (the remainder is 0, so 2 is a divisor of 5398)
5398 / 3 = 1799.3333333333 (the remainder is 1, so 3 is not a divisor of 5398)
...
5398 / 5397 = 1.000185288123 (the remainder is 1, so 5397 is not a divisor of 5398)
5398 / 5398 = 1 (the remainder is 0, so 5398 is a divisor of 5398)

Improved algorithm using square-root

However, there is another slightly better approach that reduces the number of iterations by testing only integers less than or equal to the square root of 5398 (i.e. 73.471082746888). Indeed, if a number N has a divisor D greater than its square root, then there is necessarily a smaller divisor d such that:

$D \times d = N$

(thus, if $\frac{N}{D} = d$ , then $\frac{N}{d} = D$ )

5398 / 1 = 5398 (the remainder is 0, so 1 and 5398 are divisors of 5398)
5398 / 2 = 2699 (the remainder is 0, so 2 and 2699 are divisors of 5398)
5398 / 3 = 1799.3333333333 (the remainder is 1, so 3 is not a divisor of 5398)
...
5398 / 72 = 74.972222222222 (the remainder is 70, so 72 is not a divisor of 5398)
5398 / 73 = 73.945205479452 (the remainder is 69, so 73 is not a divisor of 5398)